1.DI KETAHUI X1 Dan X2 ADALAH AKAR PERSAMAAN X2-2X+6=0 TENTUKAN PERSAMAAN KUADRAT BARU YANG AKARNYA (X1-3)DAN(X2-3)

Lanjutan

x3+x4 = x1 - 3 +x2 - 3
= x1+x2-6
= -4

x3.x4 = (x1-3)(x2-3)
=x1.x2 - 3x1 - 3x2 + 9
= x1.x2 - 3(x1+x2) + 9
= 6-3(2)+9
= 9

Maka PKB
x^2 + 4x + 9 = 0

jawab

x² - 2x + 6 = 0
x1 + x2 = 2
x1 . x2 = 6

p = x1 - 3
q = x2 - 3
p+q = (x1+x2) - 6
p+q = 2-6 = - 4

pq = (x1- 3)(x2 - 3)
pq  = (x1 x2) - 3(x1 +X2) + 9 = 6 - 3( 2) + 9
pq = 9

PK baru  x² - (p+q) x + (p q) = 0
x² - 4x + 9  - 0